0.974 013 318 541 637 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.974 013 318 541 637(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
0.974 013 318 541 637(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.974 013 318 541 637.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.974 013 318 541 637 × 2 = 1 + 0.948 026 637 083 274;
  • 2) 0.948 026 637 083 274 × 2 = 1 + 0.896 053 274 166 548;
  • 3) 0.896 053 274 166 548 × 2 = 1 + 0.792 106 548 333 096;
  • 4) 0.792 106 548 333 096 × 2 = 1 + 0.584 213 096 666 192;
  • 5) 0.584 213 096 666 192 × 2 = 1 + 0.168 426 193 332 384;
  • 6) 0.168 426 193 332 384 × 2 = 0 + 0.336 852 386 664 768;
  • 7) 0.336 852 386 664 768 × 2 = 0 + 0.673 704 773 329 536;
  • 8) 0.673 704 773 329 536 × 2 = 1 + 0.347 409 546 659 072;
  • 9) 0.347 409 546 659 072 × 2 = 0 + 0.694 819 093 318 144;
  • 10) 0.694 819 093 318 144 × 2 = 1 + 0.389 638 186 636 288;
  • 11) 0.389 638 186 636 288 × 2 = 0 + 0.779 276 373 272 576;
  • 12) 0.779 276 373 272 576 × 2 = 1 + 0.558 552 746 545 152;
  • 13) 0.558 552 746 545 152 × 2 = 1 + 0.117 105 493 090 304;
  • 14) 0.117 105 493 090 304 × 2 = 0 + 0.234 210 986 180 608;
  • 15) 0.234 210 986 180 608 × 2 = 0 + 0.468 421 972 361 216;
  • 16) 0.468 421 972 361 216 × 2 = 0 + 0.936 843 944 722 432;
  • 17) 0.936 843 944 722 432 × 2 = 1 + 0.873 687 889 444 864;
  • 18) 0.873 687 889 444 864 × 2 = 1 + 0.747 375 778 889 728;
  • 19) 0.747 375 778 889 728 × 2 = 1 + 0.494 751 557 779 456;
  • 20) 0.494 751 557 779 456 × 2 = 0 + 0.989 503 115 558 912;
  • 21) 0.989 503 115 558 912 × 2 = 1 + 0.979 006 231 117 824;
  • 22) 0.979 006 231 117 824 × 2 = 1 + 0.958 012 462 235 648;
  • 23) 0.958 012 462 235 648 × 2 = 1 + 0.916 024 924 471 296;
  • 24) 0.916 024 924 471 296 × 2 = 1 + 0.832 049 848 942 592;
  • 25) 0.832 049 848 942 592 × 2 = 1 + 0.664 099 697 885 184;
  • 26) 0.664 099 697 885 184 × 2 = 1 + 0.328 199 395 770 368;
  • 27) 0.328 199 395 770 368 × 2 = 0 + 0.656 398 791 540 736;
  • 28) 0.656 398 791 540 736 × 2 = 1 + 0.312 797 583 081 472;
  • 29) 0.312 797 583 081 472 × 2 = 0 + 0.625 595 166 162 944;
  • 30) 0.625 595 166 162 944 × 2 = 1 + 0.251 190 332 325 888;
  • 31) 0.251 190 332 325 888 × 2 = 0 + 0.502 380 664 651 776;
  • 32) 0.502 380 664 651 776 × 2 = 1 + 0.004 761 329 303 552;
  • 33) 0.004 761 329 303 552 × 2 = 0 + 0.009 522 658 607 104;
  • 34) 0.009 522 658 607 104 × 2 = 0 + 0.019 045 317 214 208;
  • 35) 0.019 045 317 214 208 × 2 = 0 + 0.038 090 634 428 416;
  • 36) 0.038 090 634 428 416 × 2 = 0 + 0.076 181 268 856 832;
  • 37) 0.076 181 268 856 832 × 2 = 0 + 0.152 362 537 713 664;
  • 38) 0.152 362 537 713 664 × 2 = 0 + 0.304 725 075 427 328;
  • 39) 0.304 725 075 427 328 × 2 = 0 + 0.609 450 150 854 656;
  • 40) 0.609 450 150 854 656 × 2 = 1 + 0.218 900 301 709 312;
  • 41) 0.218 900 301 709 312 × 2 = 0 + 0.437 800 603 418 624;
  • 42) 0.437 800 603 418 624 × 2 = 0 + 0.875 601 206 837 248;
  • 43) 0.875 601 206 837 248 × 2 = 1 + 0.751 202 413 674 496;
  • 44) 0.751 202 413 674 496 × 2 = 1 + 0.502 404 827 348 992;
  • 45) 0.502 404 827 348 992 × 2 = 1 + 0.004 809 654 697 984;
  • 46) 0.004 809 654 697 984 × 2 = 0 + 0.009 619 309 395 968;
  • 47) 0.009 619 309 395 968 × 2 = 0 + 0.019 238 618 791 936;
  • 48) 0.019 238 618 791 936 × 2 = 0 + 0.038 477 237 583 872;
  • 49) 0.038 477 237 583 872 × 2 = 0 + 0.076 954 475 167 744;
  • 50) 0.076 954 475 167 744 × 2 = 0 + 0.153 908 950 335 488;
  • 51) 0.153 908 950 335 488 × 2 = 0 + 0.307 817 900 670 976;
  • 52) 0.307 817 900 670 976 × 2 = 0 + 0.615 635 801 341 952;
  • 53) 0.615 635 801 341 952 × 2 = 1 + 0.231 271 602 683 904;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.974 013 318 541 637(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0011 1000 0000 1(2)

5. Positive number before normalization:

0.974 013 318 541 637(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0011 1000 0000 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.974 013 318 541 637(10) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0011 1000 0000 1(2) =

0.1111 1001 0101 1000 1110 1111 1101 0101 0000 0001 0011 1000 0000 1(2) × 20 =

1.1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 0111 0000 0001(2) × 2-1


7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): -1

Mantissa (not normalized):
1.1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 0111 0000 0001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-1 + 2(11-1) - 1 =

(-1 + 1 023)(10) =

1 022(10)


9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1022(10) =

011 1111 1110(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 0111 0000 0001 =

1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 0111 0000 0001


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 1110

Mantissa (52 bits) =
1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 0111 0000 0001


Decimal number 0.974 013 318 541 637 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 1111 0010 1011 0001 1101 1111 1010 1010 0000 0010 0111 0000 0001

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100